SCHAUMỖS OUTLINE SERIES THEORY AND PROBLEMS OF FOURIER ANALYSIS with applications to BOUNDARY VALUE PROBLEMS MURRAY R $PIEBEL INCLUDING 205 SOLVED PROBLEMS
SCHAUM'S OUTLINE SERIES
Trang 2SCHAUMỖS OUTLINE OF THEORY AND PROBLEMS of FOURIER ANALYSIS with Applications to Boundary Value Problems by MURRAY R SPIEGEL, Ph.D Former Professor and Chairman of Mathematics Rensselaer Polytechnic Institute of Connecticut ti
SCHAUMỖS OUTLINE SERIES McGRAW-HILL BOOK COMPANY
Trang 3Copyright ẹ 1974 by McGraw-Hill, Inc All rights reserved Printed in the United States of America No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher 0-07-060219-0 3456789101112131415 SH SH 79876 Library of Congress Catalog Card Spiegel, Murray R SchaumỖs outline of theory and problems of Fourier analysis
(SchaumỖs outline series)
1 Fourier analysis 2 Boundary value problems I Title II Title: Theory and problems of Fourier analysis
Trang 4Preface
In the early years of the 19th century the French mathematician J B J Fourier in his researches on heat conduction was led to the remarkable discovery of certain trigonometric series which now bear his name Since that time Fourier series, and generalizations to Fourier integrals and orthogonal series, have become an essential _ part of the background of scientists, engineers and mathematicians from both an ap-
plied and theoretical point of view
The purpose of this book is to present the fundamental concepts and applications of Fourier series, Fourier integrals and orthogonal functions (Bessel, Legendre, Hermite,
and Laguerre functions, as well as others)
The book is designed to be used either as a textbook for a formal course in Fourier Analysis or as a comprehensive supplement to all current standard texts It should be of considerable value to those taking courses in engineering, science or mathematics in which these important methods are frequently used It should also prove useful as a book of reference to research workers employing Fourier methods or to those inter-
ested in the field for self-study
Each chapter begins with a clear statement of pertinent definitions, principles and theorems, together with illustrative and other descriptive material The solved prob- lems serve to illustrate and amplify the theory and to provide the repetition of basic principles so vital to effective learning Numerous proofs of theorems and derivations of formulas are included among the solved problems The large number of supple- mentary problems with answers serve as a complete review of the material of each chapter
Considerably more material has been included here than can be covered in most first courses This has been done to make the book more flexible, to provide a more
useful book of reference, and to stimulate further interest in the topics
I wish to take this opportunity to thank Henry Hayden and David Beckwith for their splendid cooperation
Trang 5Chapter
CONTENTS
taining to Partial Differential Equations Linear Partial Differential Equa- tions Some Important Partial Differential Equations The Laplacian in Differ- ent Coordinate Systems Methods of Solving Boundary Value Problems
Chapter FOURIER SERIES AND APPLICATIONS
The Need for Fourier Series Periodic Functions Piecewise Continuous Fune- tions Definition of Fourier Series Dirichlet Conditions Odd and Even Functions Half-Range Fourier Sine or Cosine Series ParsevalỖs Identity Uniform Convergence Integration and Differentiation of Fourier Series, Com- plex Notation for Fourier Series Double Fourier Series Applications of Fourier Series
20
Chapter ORTHOGONAL FUNCTIONS
Definitions Involving Orthogonal Functions Orthonormal Sets Orthogonality with Respect to a Weight Function Expansion of Functions in Orthonormal Series Approximations in the Least-Squares Sense ParsevalỖs Identity for Orthonormal Series Completeness Sturm-Liouville Systems Eigenvalues and Higenfunctions The Gram-Schmidt Orthonormalization Process Applications
to Boundary Value Problems
52
Chapter GAMMA, BETA AND OTHER SPECIAL FUNCTIONS
Special Functions The Gamma Function Table of Values and Graph of the Gamma Function Asymptotic Formula for I'(n) Miscellaneous Results Involv- ing the Gamma Function The Beta Function Other Special Functions
Asymptotic Series or Expansions
67
Chapter FOURIER INTEGRALS AND APPLICATIONS
The Need for Fourier Integrals The Fourier Integral Equivalent Forms of FourierỖs Integral Theorem Fourier Transforms Fourier Sine and Cosine Transforms ParsevalỖs Identities for Fourier Integrals The Convolution Theorem for Fourier Transforms Applications of Fourier Integrals and Trans-
forms
80
Chapter BESSEL FUNCTIONS AND APPLICATIONS
BesselỖs Differential Equation The Method of Frobenius Bessel Functions of the First Kind Bessel Functions of the Second Kind Generating Function for J,(%) Recurrence Formulas Functions Related to Bessel Functions Equa- tions Transformable into BesselỖs Equation Asymptotic Formulas for Bessel
Functions Zeros of Bessel Functions Orthogonality of Bessel Functions of the
First Kind Series of Bessel Functions of the First Kind Orthogonality and Series of Bessel Functions of the Second Kind Solutions to Boundary Value Problems Using Bessel Functions
Trang 6CONTENTS Page
Chapter Z7 LEGENDRE FUNCTIONS AND APPLICATIONS 130
LegendreỖs Differential Equation Legendre Polynomials Generating Function for Legendre Polynomials Recurrence Formulas Legendre Functions of the
Second Kind Orthogonality of Legendre Polynomials Series of Legendre
Polynomials Associated Legendre Functions Orthogonality of Associated Legendre Functions Solutions to Boundary Value Problems Using Legendre Functions *
Chaptr ậ& HERMITE, LAGUERRE
AND OTHER ORTHOGONAL POLYNOMIALS 154
HermiteỖs Differential Equation Hermite Polynomials Generating Function for Hermite Polynomials Recurrence Formulas for Hermite Polynomials Orthogonality of Hermite Polynomials Series of Hermite Polynomials La- guerreỖs Differential Equation Laguerre Polynomials Some Important Prop- erties of Laguerre Polynomials, Miscellaneous Orthogonal Polynomials and Their Properties
Appendix A Uniqueness of Solutions ề0.0.0.0 000s cee cece cece eee eencececuteunes 167 Appendix B Special Fourier Series 0.0 ccc cece eee eee eet ence neeeunes 169
Appendix C Special Fourier Transforms .0 0 00.0 c ete cc cece ceuuecvcueureans 173
Appendix D Tables of Values for Jo(x) and Jy(@) 6 ccc cc ce cece cece eeenes 176
Appendix EJ = Zeros of Bessel Functions .00ccccceeeeecccucccecunececeueeeees 177
Trang 7Chapter 1
Boundary Value Problems
MATHEMATICAL FORMULATION AND SOLUTION OF PHYSICAL PROBLEMS
In solving problems of science and engineering the following steps are generally taken 1 Mathematical formulation To achieve such formulation we usually adopt mathematical
models which serve to approximate the real objects under investigation
Example 1
To investigate the motion of the earth or other planet about the sun we can choose points as mathe- matical models of the sun and earth On the other hand, if we wish to investigate the motion of the earth about its axis, the mathematical model cannot be a point but might be a sphere or even more accu- rately an ellipsoid
In the mathematical formulation we use known physical laws to set up equations describing the problem If the laws are unknown we may even be led to set up experi-
ments in order to discover them
Example 2
In describing the motion of a planet about the sun we use NewtonỖs laws to arrive at a differential equation involving the distance of the planet from the sun at any time
2 Mathematical solution Once a problem has been successfully formulated in terms of
equations, we need to solve them for the unknowns involved, subject to the various
conditions which are given or implied in the physical problem One important con- sideration is whether such solutions actually exist and, if they do exist, whether they are unique
In the attempt to find solutions, the need for new kinds of mathematical analysis Ở leading to new mathematical problems Ở may arise
Example 3
J.B.J Fourier, in attempting to solve a problem in heat flow which he had formulated in terms of partial differential equations, was led to the mathematical problem of expansion of functions into series involving sines and cosines Such series, now called Fourier series, are of interest from the point of view of mathematical theory and in physical applications, as we shall see in Chapter 2
3 Physical interpretation After a solution has been obtained, it is useful to interpret it physically Such interpretations may be of value in suggesting other kinds of problems, which could lead to new knowledge of a mathematical or physical nature
In this book we shall be mainly concerned with the mathematical formulation of physi-
Trang 82 BOUNDARY VALUE PROBLEMS (CHAP 1
*
DEFINITIONS PERTAINING TO PARTLAL DIFFERENTIAL EQUATIONS
A partial differential equation is an equation containing an unknown function of two or more variables and its partial derivatives with respect to these variables
The order of a partial differential equation is the order of the highest derivative present
Example 4
82%
Ox dy
equation Here u is the dependent variable while x and y are independent variables
A solution of a partial differential equation is any function which satisfies the equation identically
The general solution is a solution which contains a number of arbitrary independent functions equai to the order of the equation
= 2xỞy is a partial differential equation of order two, or a second-order partial differential
A particular solution is one which can be obtained from the general solution by particu- lar choice of the arbitrary functions
Example 5
As seen by substitution, u = x?y Ở Jay?+ F(x)+G(y) is a solution of the partial differential equation of Example 4, Because it contains two arbitrary independent functions F(aỢ) and G(y), it is the general solution If in particular F(x) = 2sina, G(y) = 8y4Ở5, we obtain the particular solution
u = xy Ở duy? + 2sina + 3By4 Ở 5
A singular solution is one which cannot be obtained from the general solution by par- ticular choice of the arbitrary functions
Example 6
If w= eae oe)? where ề is a function of x and y, we see by substitution that both
u = ềxF(y)Ở [F(y)|? and u = 22/4 are solutions The first is the general solution involving one arbitrary function F(y) The second, which cannot be obtained from the Ổgeneral solution by any choice of F(y),
is a singular solution
A boundary value problem involving a partial differential equation seeks all solutions of the equation which satisfy conditions called boundary conditions Theorems relating to the existence and uniqueness of such solutions are called existence and uniqueness theorems
LINEAR PARTIAL DIFFERENTIAL EQUATIONS
The general linear partial differential equation of order two in two independent vari- ables has the form
ru au neu ou ou _
where A,B, ,G may depend on ề and y but not on wu A second-order equation with
independent variables x and y which does not have the Ộorm (1) is called nonlinear
If G=0 identically the equation is called homogeneous, while if G0 it is called non- homogeneous Generalizations to higher-order equations are easily made
Because of the nature of the solutions of (1), the equation is often classified as elliptic,
hyperbolic, or parabolic according as B?Ở4AC is less than, greater than, or equal to zero,
Trang 9CHAP 1] BOUNDARY VALUE PROBLEMS 3
SOME IMPORTANT PARTIAL DIFFERENTIAL EQUATIONS 1 a Yo ply 074 Vibrating string equation ap = Cap
This equation is applicable to the small transverse vibrations of a taut, flexible string,
such as a violin string, initially located on the Ụ
x-axis and set into motion (see Fig 1-1) The
function y(a#,f) is the displacement of any
point ề of the string at time t The constant y(a, t) x a? =7/n, where 7 is the (constant) tension in |
the string and Ừ is the (constant) mass per unit length of the string It is assumed that
no external forces act on the string and that Fig 1-1
it vibrates only due to its elasticity The equation can easily be generalized to higher dimensions, as for example the vibrations of a membrane or drumhead in two dimensions ắn two dimensions, the equation is Pz _Ở ,(0% , 0z mm : : Ou Heat conduction equation aT Vu
Here w(x, y, z, t) is the temperature at position (x,y,z) in a solid at time ý, The con- stant x, called the diffusivity, is equal to R/ơu, where the thermai conduectivitụ K, the specific heat o and the density (mass per unit volume) Ừ are assumed constant We call Vu the Laplacian of u; it is given in three-dimensional rectangular coordinates X,Y, 2) by ( ) G3, ou Oru da? ay? 0e Vu LaplaceỖs equation V?u = 0
This equation occurs in many fields In the theory of heat conduction, for example, v is the steady-state temperature, ie the temperature after a long time has elapsed, whose equation is obtained by putting du/dat =0 in the heat conduction equation above In the theory of gravitation or electricity v represents the gravitational or electric potential respectively For this reason the equation is often called the potential equation The problem of solving V?v =0 inside a region R when v is some given function on the boundary of ệ is often called a Dirichlet problem ru, Ou Longitudinal vibrations of a beam 3B 7 Oop
This equation describes the motion of a beam (Fig 1-2, page 4) which can vibrate longitudinally (i.e in the x-direction) the vibrations being assumed small The variable u(x, t) is the longitudinal displacement from the equilibrium position of the cross section at + The constant ằ? = E/n, where E is the modulus of elasticity (stress divided by strain) and depends on the properties of the beam, Ừ is the density (mass per unit
volume)
Trang 10SA BOUNDARY VALUE PROBLEMS Ẽ _ IOHAP.I 5 Transverse vibrations of a beam = + BPS
This equation describes the motion of a beam (initially located on the x-axis, see
Fig 1-8) which is vibrating transversely (ie perpendicular to the z-direction) assuming small vibrations In this case y(x,t) is the transverse displacement or deflection at any time ằt of any point x The constant b? = EI/Au, where E is the modulus of elasticity, I is the moment of inertia of any cross section about the x-axis, A is the area of cross section and Ừ is the mass per unit length In case an external transverse force F(z, t) is applied, the right-hand side of the equation is replaced by b?F(#, t)/EI
,
Fig.1-2 Fig.1-3
_THE LAPLACIAN IN DIFFERENT COORDINATE SYSTEMS
The Laplacian vu often arises in partial differential equations of science and engi- neering Depending on the type of problem involved, the choice of coordinate system may be important in obtaining solutions For example, if the problem involves a cylinder, it: Ổ will often be convenient to use cylindrical coordinates; while if it involves a sphere, it will: _ be convenient to use spherical coordinates
The Laplacian i in cylindrical coordinates (p, #, 2): (see Fig 1-4) is given by
au lau law aw
2, -_ 2" a Ở -
Vv t - OpỢ (pop + p= dd + az? Ấ (?)
The transformation equations between rectangular and cylindrical coordinates are
= peg, y= psing, 2=2
where p= 0, 0SƯ<2z, Ở~<2.< @, ,
The Laplacian i in spherical coordinates ứ, 6,) (see Fig 1-5) is given by
Trang 11CHAP 1] BOUNDARY VALUE PROBLEMS 5
1 oa ou 1 af du 1 8ồ
9 Ở Ở _Ở 2 Ở ỞỞ ỞỞỞ _Ở
VM= 7? or ứ | + r? sin 6.06 (sin a) + r sin? 6 0ằ? : (4) The transformation equations between rectangular and spherical coordinates are
x = rsinécosằ, y = rsinésing, z = rcosé (5) where r=0, OS 057, 05 ằ<2r
METHODS OF SOLVING BOUNDARY VALUE PROBLEMS
There are many methods by which boundary value problems involving linear partial differential equations can be solved In this book we shall be concerned with two methods which represent somewhat opposing points of view
In the first method we seek to find the general solution of the partial differential equa- tion and then particularize it to obtain the actual solution by using the boundary condi- tions In the second method we first find particular solutions of the partial differential equation and then build up the actual solution by use of these particular solutions Of the two methods the second will be found to be of far greater applicability than the first 1 General solutions In this method we first find the general solution and then that par-
ticular solution which satisfies the boundary conditions The following theorems are of fundamental importance
Theorem 1-1 (Superposition principle): If 1:1, %2, .,%, are solutions of a linear ho- mogeneous partial differential equation, then Ạ1i -} eswa + - + Can, where ằ1, ằ2, ,ằn are constants, is also a solution
Theorem 1-2: The general solution of a linear nonhomogeneous partial differential equa- tion is obtained by adding a particular solution of the nonhomogeneous equation to the general solution of the homogeneous equation
We can sometimes find general solutions by using the methods of ordinary differen- tial equations See Problems 1.15 and 1.16
If A,B, ,F in (1) are constants, then the general solution of the homogeneous equation can be found by assuming that u= esttby) where a and 0 are constants to be determined See Problems 1.17~1.20
Trang 12BOUNDARY VALUE PROBLEMS [CHAP 1 Solved Problems MATHEMATICAL FORMULATION OF PHYSICAL PROBLEMS 1.1 1.2 1.3 Derive the vibrating string equation on page 3
Referring to Fig 1-6, assume that As represents Ti
an element of arc of the string Since the tension is Soy 72 I
assumed constant, the net upward vertical force acting 7 cs y on As is given by | 4 T sin 6 Ở 7 sin 6; (2) a | : : _ : i | ; Since sin 6 = tané, approximately, for small angles, x e+ Ae this force is ay _Ở Ấ9 Fig 1-6 "8a x+Az Tạ x (2) eh : : _ 0 9 ey using the fact that the slope is tang = =~ We use here the notation Ở~| and Ở* for the Ox 0% {x Ow |at+Ax
partial derivatives of y with respect to x evaluated at ề and x + Az, respectively By NewtonỖs law
this net force is equal to the mass of the string (u As) times the acceleration of As, which is given by 2 oy + e where ề>0 as As>0 82 Thus we have approximately | = (sa (TH + ề) (8) If the vibrations are small, then As = Ax approximately, so that (3) becomes on division by Az: oy _ "| ax x+Az Ox oy _ 9 7 OX atAzr Ox |e = 03 nde Ở aặ toe (4) - Taking the limit as Az > 0 (in which case ề > 0 also), we have \ 2 2 2 ate = Fy yo My _ iy
ỘOx ax ms at ae = Ở gaz? Where a? = r/n 2 Ở=
Write the boundary conditions for a vibrating string of length L for which (a) the ends +=0 and x=L are fixed, (b) the initial shape is given by f(x), (c) the initial
velocity distribution is given by g(x), (d) the displacement at any point x at time t is bounded
(a) If the string is fixed at x= 0 and ề=L, then the displacement y(ề,t) at ề= 0 and ề=L must be zero for all times t > 0, ie y(0,t) = 0, y(L,t) = 0 t>0 (b) -Since the string has an initial shape given by f(x), we must have (z,0) = f(x) O<a<L (ec) Since the initial velocity of the string at any point x is Ổ9(x), we must have y(xz,0) = g(x) O<a<L
Note that y,(z, 0) is the same as dy/dt evaluated at ặ = 0
(đ) Since y(a, t) is bounded, we can find a constant M independent of x and t such that
|u(Ủ,t)| < M 0<z<1L, t>0
Write boundary conditions for a vibrating string for which (a) the end ề=0 is moving so that its displacement is given in terms of time by G(t), (0) the end z = 7,
is not fixed but is free to move
(a) The displacement at += 0 is given by y(0,t) Thus we have
Trang 13CHAP 1] BOUNDARY VALUE PROBLEMS 1 (6) If 7 is the tension, the transverse force acting at any point x is 9 Tan = ry,(a, t) Since the end ề=L is free to move so that there is no force acting on it, the boundary condition is given by TyầtL,t) = 0 or y,(Z,t) = 0 t>0
14 Suppose that in Problem 1.1 the tension in the string.is variable, i.e depends on the
particular point taken Denoting this tension by (x), show that the equation for the vibrating string is ở | _ 02 sal aa] = tạp In this case we write (2) of Problem 1.1 as oy _ oy r() Ox |x+ Ax 7(2) Ox [x so that the corresponding equation (4) is oy Ở oy r(x) ô# |z+A+ 7Ủ) Ou |x 02w = Ởz + < pA at? Thus, taking the limit as Ax > 0 (in which case ề> 0), we obtain 2| Ấ9| _ Ấ9ồ de [e2 ae | Ở Ộấp after multiplying by ụ
1.5 Show that the heat flux across a plane in a conducting medium is given by ỞK ot, where w is the temperature, n is a normal in a direction perpendicular to the plane and K is the thermal conductivity of the medium
Suppose we have two parallel planes I and II a dis- I II tance An apart (Fig 1-7), having temperatures u and
u-+ Au, respectively Then the heat flows from the plane of higher temperature to the plane of lower temperature Also, the amount of heat per unit area per unit time, called the heat flux, is directly proportional to the difference in temperature Au and inversely proportional to the distance An Thus we have
1 ut Au An
Heat fluxfromItoII = Ke (7) where K is the constant of proportionality, called the ther-
mal conductivity The minus sign occurs in (1) since if
Au > 0 the heat flow actually /takes place from II to I Fig 1-7 By taking the limit of (7) as An and thus Au approaches zero, we have as required:
Heat flux across planeI = -K (2) We sometimes call a the gradient of u which in vector form is Vu, so that (2) can be written
Heat flux across planeI = ỞK VựẤ (8)
16 If the temperature at any point (x,y,z) of a solid at time ằ is u(x, y,2,t) and if K,o
and Ừ are respectively the thermal conductivity, specific heat and density of the solid,
Trang 14BOUNDARY VALUE PROBLEMS [CHAP 1
a = ềVu where ề = /ơu
Consider a small volume element of the solid V, as indicated in Fig 1-8 and greatly enlarged
in Fig 1-9 By Problem 1.5 the amount of heat per unit area per unit time entering the element
3% ou
through face PQRS is ỞK aul, where az |x
ated at the position x Since the area of face PQRS is Ay Az, the total amount of heat entering the element through face PQRS in time At is
indicates the derivative of u with respect to x evalu- du ỞK Ka [Ay acat (1) 1 Similarly, the amount of heat leaving the element through face NWZT is 91 ỞKẠ~ 2 Ơ# Ìz+Az Ay Az At (2) where = ta indicates the derivative of u with respect to ề evaluated at 2+ Az x +
The amount of heat which remains in the element is given by the amount entering minus the amount leaving, which is, from (1) and (2),
ou \k 9#
In a similar way we can show that the amounts of heat remaining in the element due to heat transfer taking place in the y- and z-directions are given by Ở gân vai (3) x z+Az 9# du ou {x ay lytay Kay Lj} Aw Az At (4) 0% ou and {ề 32 z+Az ae + Aa Ay At (5) respectively
Trang 15CHAP.I Ẽ ể - BOUNDARY VALUE PROBLEMS ` M,
Fig.1-8 Ở Fig.I-9 1.7 Work Problem 1.6 by using vector methods
Let V be an arbitrary volume lying within the solid, and let S denote its: surface (see Fig 1-8) The total flux of heat across S, or the quantity of heat leaving S: per unit time, is _ SJ x90) mas " " where n is an outward-drawn unit normal to S Thus the quantity of heat entering S.per unit time isỢ Ặ (Wu)*ndứ = ff V+(KVu) dv (1) 8 Vv : Ổby the divergence theorem The heat contained in a volume V is given by SSf onudV er Vv : : Ổ Then the time rate of increase of heat is ml se wi SJ : oped = JlẶss = oH p dV (2) a) V =: / Equating | the right-hand sides of (1) ) and (2 ),
SSS [z% Tên Ộeve | ave = 0
and since V is arbitrary, the: integrand, assumed | continuous, must be identically zero, so that ` bu ST s1 so ose = V*(Wu) or if K,o,Ừ are constants, Ở an : Ở ou = Aw K = 2 aE oa Ve Vu | nV24, (8)
1.8 Show that for steady-state heat flow the heat conduction equation of Problem 1.6 or Ừ LT reduces to LaplaceỖs equation, 2w = 0
In the case of steady-state heat flow the temperature u does not depend on time t, so Ừ that a
3? = 0 Thus the equation oe = ềV2u becomes V2u = 0
Trang 1610 BOUNDARY VALUE PROBLEMS [CHAP 1
(a) If the initial temperature is f(x) and the ends are kept at temperature zero, set up
the boundary value problem (b) Work part (a) if theend ề= L is insulated (c) Work part (a) if the end ề=L radiates into the surrounding medium, which is assumed to
be at temperature wo
This is a problem in one-dimensional heat con-
duction since the temperature can only depend on # the position x at any time ằ and can thus be de-
noted by u(#,t) The heat conduction equation is thus given by ou 2a Ở 2 = ka Ở O<a<L,t>0 > Ể) 1 Fig 1-10 g (a) Since the ends are kept at temperature zero, we have (0,0) = 0, u(L,t) = 0 t>0 (2) Since the initial temperature is f(x), we have w(Ủ,0) = f(x) O<a<L (3) Also, from physical considerations the temperature must be bounded; hence ju(x, t)| < M 0<z<L, t>0 (4) The problem of solving (1) subject to conditions (2), (3) and (4) is the required boundary value problem A problem exactly equivalent to that considered above is that of an infinite slab of conducting material bounded by the planes ề= 0 and x= UL, where the planes are kept at temperature zero and where the temperature distribution initially is f(x)
(b) If the end ề=L is insulated instead of being at temperature zero, then we must find a
replacement for the condition u(L,t)=0 in (2) To do this we note that if the end x=L is insulated then the flux at ề Ở L is zero Thus we have
31 Ở : _
ỞK = 0 or equivalently wu,(L,t) = 0 (5) which is the required boundary condition
(c) It is known from physical laws of heat transfer that the heat flux of radiation from one object at temperature U, to another object at temperature U, is given by a(Ut Ở US), where a is
a constant and the temperatures U, and Uy, are given in absolute or Kelvin temperature which
ig the number of Celsius (centigrade) degrees plus 273 This law is often ealled StefanỖs
radiation law From this we obtain the boundary condition
ỞKu,(L,t) = aluiỞ ul) where u, = u(L, t) (6) If and wy do not differ too greatly from each other, we can write
4 4 _ 3 2 2
tị Up = (Uy Ở Up)(Uy + UjUy + Uy + or)
3
- (uy word (2) + (By +5 +Ở + 1]
= Aug (uy Ở Up)
II
since (Ộ4/,g)3, (wy/wạ)2, (uy/u9) are approximately cqual to 1, Using this approximation, which is often referred to as NewtonỖs law of cooling, we can write (6) as
ỞKu,(L,t) = B(uy Ở mạ) (7)
where # is a constant
CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS
Trang 17CHAP 1] BOUNDARY VALUE PROBLEMS ồ< 11 ou aru Ở_ = Ở linear, order 2, dep var u, ind var x,t (0) 3 ja? ồ Ộ 8 2 () ụ? a = yok linear, order 3, dep var R, ind var x, y ew (c) Woe = rst , nonlinear, order 2, dep var W, ind var 7,8, t 2 2 2 (a 2% 42% 4% = 9 Sa? ay? PP linear, order 2, dep var ằ, ind var #, , 2 (e Ở 2 + oz + 3z ồ = |] nonlinear, order 1, dep var z, ind var u,v OU ov 1.11 Classify each of the following equations as elliptic, hyperbolic or parabolic ep , HH _ (a) 022 oy? = 0 u=ằ, A=1, B=0, C=1; B*-Ở4AC=-4<0 and the equation is elliptic 0w su (0) ủy = xu >Ở=t, A=xk, B=0, C=0; B2?Ở4AC=0 and the equation is parabolic ry _ Ấ0 (c) ot? = a on? y=t, u=y, A=a*, B=0, C=-1; B*-Ở4AC=4a2>0 and the equation is hy- perbolie anu 0? 0?w, au au Ở> Ở> =ỞỞỞ 2ỞỞ Âu = 2z Ở 5 (d) 2a + Banay + đạn Tống 2a + % +Ấ A=1, B=8, C=4; B?-Ở4AC=-Ở7<0 and the equation is elliptic PU 0ệ Ou
A=a, B=0, C=y; B*-Ở4AC =-Ở4zy Hence, in the region xy > 0 the equation is elliptic; in the region xy <0 the equation is hyperbolic; if xy = 0, the equation is parabolic
SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS
1.12 Show that u(a#,t) =e sin2z is a solution to the boundary value problem au Oru
at Ở 2a? u(0,t) = u(x,t) = 0, u(x,0) = sin 2x
From u(x,t) = eỞ8t sin2xỢ we have
u(0,t) = e-8t sind = 0, u(z,t) = e-8tsin2z = 0, uf#,0) = e-%sin2e = sin 2x
and the boundary conditions are satisfied Ou
Also at 2
= Ở8e-ật sìn 2z, du 2eỞ8t cos 2x, 3# Ở = Ở4eỞ8t gin 2x 9# Then substituting into the differential equation, we have
Trang 1812 1.13 (a) (0) (a) (b) 114 (a) (0) {a) ()
BOUNDARY VALUE PROBLEMS [CHAP 1
Show that ò=#(yỞ3z), where #' is an arbitrary diferentiable function, is a
general solution of the equation
ov Ov
ax + 3a = 0
Find the particular solution which satisfies the condition v(0,y) = 4siny
+
Let yỞ3x2 =u Then v=F(u) and
ov pm YOU Fw ae FỖ(u(Ở-3) = NoN\t Ở _ am Ở8F"(u)
av _ ôu ữw , =_ gt ây bu ay FỖ(u)(1) F'(u)
av ov
Thus aa + oy 0
Since the equation is of order one, the solution v = F(u) = F(yỞ 3x), which involves one arbitrary function, is a general solution -
v(a,y) = F(yỞ8x) Then v(0,y) = Fly) = 4siny But if Fly) = 4 sin y, then v(z,y) = F(y Ở 3z) = 4 sin (yỞ 8x) is the required solution
Show that (z,f) = F(2z + 5t) + G(2zỞ 5ặ) is a general solution of ey Ở 3?
đạp = 25p:
Find a particular solution satisfying the conditions
y(0,t) = y(z,t) = 0, ~=Ở-y(w, 0) = sin2z, y(x,0) = 0
Let 24+ 5t=u, 2aỞ5t=v Then y = F(u)+ Giv)
w= Fu, eG So = Puy) + G()Ở8) = 5E") Ở 5⁄00) (1)
Se = GOP wỞsa) = 52h! eu _ oe = BBR) + 256"(v) (2)
Se = eo out + aG oe = F'(u)(2) + G2) = 2F"(u) + 2G/(v) (8) oy = 2 [2F'(u) +26'0)] = gifỢ ou + gic = =_ 4FỢ(w) + 4G'() (4)
2 2
From (2) and (4), aoe = 25 <4 and the equation is satisfied Since the equation is of
order 2 and the solution involves two arbitrary functions, it is a general solution
Trang 19CHAP 1] BOUNDARY VALUE PROBLEMS 13 from which F(2x) = Ậsin2z + eụ, G(2+) = Jsin2x + ey
i.e ya, t) = fsin(2a+5t) + J sin (2x Ở5t) + cy + ey Using y(0,t)}=0 or y(z,t)=0, e,+ằe,=0 so that
y(z,t) = $Ậsin(2z+ỏđặ) + j sin(2zỞBU) Ở sin 2x cos 5t which can be checked as the required solution
METHODS OF FINDING SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS
2
1.15 (a) Solve the equation 0# 0 = wy
(6) Find the particular solution for which z(z,0) = 2z, z(1,) Ở cosỹ
(a) Write the equation as 2(#) = wy Then integrating with respect to x, we find dx \ay
Oz
Se _Ở Ags ay 3zồw + Fly) (1)
where Fy) is arbitrary
Integrating (1) with respect to y,
z = gay? + Ặru dy + G(x) (2)
where G(x) is arbitrary The result (2) can be written
# = Z#,U) = $z3 + Aly) + G(x) (3)
which has two arbitrary (independent) functions and is therefore a general solution (b) Since z(+,0) = z?, we have from (#)
z2 = H(0) + G(z) or G(x) = a2 Ở H(0) (4) Thus z = 4#9 + Hy) + 2 Ở H(0) (5),
Since 2z(1,y) = cosy, we have from (5)
cosy = dy? + Hly) + 1 Ở H() or Hy) = cosy Ở ty? Ở 1+ H(0) (6) Using (6) in (5), we find the required solution
Trang 2014
1.18
1.19
1.20
BOUNDARY VALUE PROBLEMS [CHAP 1
Assume u = e+by, Substituting in the given equation, we find (a2 + 38ab + 2b2)eart+by = 0 or a2? + 8ab + 2b2 = 0
Then (a+b)(a+2b)=0 and a=Ởb, a=-Ở2b If a=Ởb, ee t+by = gb@Ở#) ig a solution for any value of b If a= Ở2b, eỞ2bx+by = eb(yỞ2z) ig a solution for any value of b
Since the equation is linear and homogeneous, sums of these solutions are solutions (Theorem
1-1) For example, 3e2Ể-Ởz) TỞ 2/8@~Ởz) + Bemệ%Ởz) ig a solution (among many others), and one is thus led to F'(y Ở x) where F is arbitrary, which can be verified as a solution Similarly, G(y Ở 2a), where G is arbitrary, is a solution The general solution found by addition is then given by
u = F(@wỞ#) + G(yỞ 2z)
: ou au Pu a , d?w Ở
Find a general solution of (a) 2+ ST = 2u, (b) 42 + ây + aye 0
(a) Let wu = ew+by, Then 2ụ + 8b = 2, a= 258 and ụ[(2~3b)/2]z + by Ở exe(b/2)(2yỞ3x)
is a solution ồ
Thus u = e*F(2y Ở 3a) is a general solution
(b) Let w = esttoy, Then 4a2~Ở4ab+ 62 = 0 and 6b = 2a,2a From this u = ee+2y) and so F(a + 2y) is a solution
By analogy with repeated roots for ordinary differential equations we might be led to believe ềG(x + 2y) or yG(a+2y) to be another solution, and that this is in fact true is easy to verify Thus a general solution is + u = Flat 2y) + zG(s + 2y) or ` u = F(x +2y) + yG(u t+ 2y) ru Pu Solv ồ Ở, aa + oy? =z = 10e**Ừ, : au , Pu _ _ : The homogeneous equation 2m2 T aye = 0 has general solution wu = F(a+ iy) + G(a Ở iy) by Problem 1.39(c)
To find a particular solution of the given equation assume 1% = ae2*+ầ where ề is an unknown
constant This is the method of undetermined coefficients as in ordinary differential equations
We find a= 2, so that the required general solution is
Ộu = Flet+iy) + GlaỞiy) + 2e2r+y
au Pu nay
Solve aa? 4 ap Oe
The homogeneous equation has general solution
+ = Ƒ(23z+w) + G(2 Ở g)
To find a particular solution, we would normally assume u = ae2*+ầ as in Problem 1.19 but this assumed solution is already included in F(2x%+y) Hence we assume as in ordinary differential equations that % Ở aỦe2zẨ1 (or u = aye2**ầ), Substituting, we find ề= 1
Trang 21CHAP 1] BOUNDARY VALUE PROBLEMS 15
1.22
1.23
is a solution By the last boundary condition,
by the method of separation of variables
Let ề= XY in the given equation, where X depends only on x and Y depends only on y Then XY -= 43XY or X'/4X = Y'/Y
where XỖ = dX/dx and Y' = dY/dy
Since X depends only on x and Y depends only on y and since ề and y are independent vari-
ables, each side must be a constant, say c
Then XỖỞ4eX =0, YỖỞcY =0, whose solutions are X = Á4e1z, Y = Bec,
A solution is thus given by
ule,y) = XY = ABec4z+y = Kec(z+u)
From the boundary condition,
u(0,y) = Key = 8e~3%
which is possible if and only if K=8 and e=-Ở8 Then #(z,) Ở 8eỞ34zẨ) = 8eỞ~122Ở3w jg the required solution
Solve Problem 1.21 if u(0,y) = 8e7-* + 4ụ~5*,
As before a solution is Kec(4+Ỏ, Then K,e%14z+ and Kyec24ằ+4) are solutions and by the principle of superposition so also is their sum; i.e a solution is
u(a,y) = Kyer4etw + Kyecose+y)
From the boundary condition,
u(0,y) = Kye + KyeỖ = BeỞ3 + 4eỞ~5w
which is possible if and only if K, = 8, K,=4, ằ, = Ở3, eg = Ở5
Then u(%,y) = 8e~3(ặẨz) + 4eỞ5(4z+w) = 8eỞ12z~Ở 3u + 4e~20rỞ5y jg the required solution
au 02
Solve 3f = 2m 0<z<3, (>0, given that {0,f) = (3,0 = 0,
w(z,0) = 5sin4zz Ở 3sin8zz + 2sin l0zz, |u(z,#)| < M where the last condition states that u is bounded for 0<2<3,t> 0
Let w= XT Then XTỖ =X"T and X"/X = T'/2T Each side must be a constant, which we call Ở? (If we use +22, the resulting solution obtained does not satisfy the boundedness condi-
tion for real values of A.) Then
xXỢ + 2X = 0, T' + 2VT = 0
with solutions X = A, cosdrw + By sin Ax, T = ee 2ầt
A solution of the partial differential equation is thus given by
u(x,t) = XT = eye-2t (AieosA + BisinAz) = e72"t(A cosax + B sin da)
Since u(0,t) = 0, e~?Ợt(A)=0 or A=0 Then
u(z, t) = Be~2t sin dx
Since (8,ặ#) =0, Be~?*Ợ! sin3A =0 If B=0, the solution is identically zero, so we must have
sin3A\=0 or 3A = mr, }\ = m7z/3, where m= 0,+1,+2, Thus a solution is
Trang 2216 124 Thus (2) becomes ỘBOUNDARY VALUE PROBLEMS = -{CHAP 1 mye
ule; 0) = Bị sin-EỞ 3 + By sin Morx 3 + Bs sin + Ngrh 3 = ậ5sin4drx Ở 3 sin 8z + 2 sin 107%
which is possible if-and only if By=Ở5, m,= 12, By= Ở3, My, = 24, Bs = 2, mạ = 30 Substituting these in (2), the required solution is
ula, t) = Be 820ồ gin Am Ở 3e~128rồt sin 87% + Qe~ 2000" sin 107% (2) This boundary value problem-has the following interpretation as Ổa heat flow problem A bar
Ộwhose surface is insulated (Fig 1-11) has a length of 3 units and a diffusivity of 2 units If its ends
are kept at temperature zero units and its initial temperature, u(v,0) = 5 sin Aare Ở 8 sin 87% + 2.sin 107%, find the temperature at position x at time ằ, i.e find: u(x, t) `
Fig 1-11
Solve a = 165 y 0<2<2, t>0, subject to the conditions (0, t) = 0, y(2, t) = 0, (Ủ, 0) = 6 sin x# Ở 8 sỉn 4mz, (Ủ,0) = 0, [y(a, ệ)| <M
Let y= XT, where X depends only on x, 1 depends only on ý, Then substitution in the differential equation yields ể
XTỢ = 16X"T or X"/X = TỖ/16T
on separating the variables Since each side must be a constant, say ỞN, we have
XỢ +X = 0, T+1627 = 0
Solving these we find :
X = a,coska + by sinaw, 7 = Ge cos 4kằt + by sin 4At
Thus a solution is
y(a,t) = (a, cos À# + 6, sin dalay cos 44ằ + bs sin 4X4) (1)
To find the constants it is simpler to proceed by using: firstỢ those boundary conditions involving
two zeros, such as y(0,ằ) = 0, y,(#,0) = 0 From -y(0,t) = 0 we see from (1) that :
đ1(0Ư Ủos 4AXặ + by sin 4At) =
so that to obtain a non zero solution (7) we must have a; = 0 Thus (1) becomes
y(w,t) = (by sin Aw)(ay cos 4\t + by sin 4At) _ (2) - Differentiation of (2) with respect to ặ yields 7
yi(x, t) = - (bị sim Nx)(Ở4dag sin 4nt + 4dBe cog 4X4) Ộgo that we have on 1 putting ặ=0 and using the condition y;(ề#,0) = 0
/(@,0) = (by sin Aw)(4ab) = 0 : " s (8) : In order to obtain a solution (2) which is not zero we see from (3) Ổthat we must have by = =0
_M(, j= B sin Ax cos 5 Ant
on 2 putting by = 0- and writing B= Pada
From vớ, t) = 0 we now find :
Bsin2^A cos 4^Ư = 0
Trang 23CHAP 1] BOUNDARY VALUE PROBLEMS 17
1.25
1tr2
2 008 2mrt (4)
Thus y(x,t) = Begin
is a solution Since this solution is bounded, the condition |y(z, 9| < M is automatically satisfied In order to satisfy the last condition, y(z,0) = 5 sinzz Ở 3 sin4zx, we first use the principle
of superposition to obtain the solution
Tham+%
,_ Myre 2
y(x,t) = By, sin cos 2my7t + By sin gỞ C08 2Mort (5) Then putting tỞ0 we arrive at _ Myre _ More (z,0) = By, sin _ + Bz sin 5 = 6sin 7x Ở 3 sin 4r% This is possible if and only if B,=6, m,=2, B,=-Ở8, m,g=8 Thus the required solution (5) is
(z,f) =_ 6 sinze cos 4rt Ở 8 sin 47x cos 16rt (6) This boundary value problem can be interpreted physically in terms of the vibrations of a string The string has its ends fixed at ề=0 and ề =2 and is given an initial shape f(x) = 6 sỉn z Ở 3 sin 4zz It is then released so that its initial velocity is zero Then (6) gives the displacement of any point x of the string at any later time ý
2
Solve à = 2, 0<z<3, t>0, given that u(0,t)=u(3,t)=0, u(x,0) = f(a),
|Ấ(Ủ, 9| < M
This problem differs from Problem 1.23 only in the condition z{z,0) =/() In seeking to satisfy this last condition we see that taking a finite number of terms, as in (1) of Problem 1.23, will be insufficient for arbitrary f(x) Thus we are led to assume that infinitely many terms are taken, i.e
~ ;Ấ TH
w(Ủ,Đ = met ứ BmeỞ-2m r9 sịn TCỢ 3
The condition u(x,0) = f(x) then leads to
fe) = & BysinỎZ ma1
or the problem of expansion of a function into a sine series Such trigonometric expansions, or
Fourier series, will be considered in detail in the next chapter
Supplementary Problems
MATHEMATICAL FORMULATION OF PHYSICAL PROBLEMS 1.26
127
If a taut, horizontal string with fixed ends vibrates in a vertical plane under the influence of gray-
ity, show that its equation is
đ5V _Ở Ấ;đồw
at? 8z2
where ụ is the acceleration due to gravity
A thin bar located on the z-axis has its ends at z0 and z= The initial temperature of the bar is /), 0< ặ << L, and the ends ề =0,2=L are maintained at constant temperatures T,, T, respectively Assuming the surrounding medium is at temperature Ug and that NewtonỖs law of cooling applies, show that the partial differential equation for the temperature of the bar at any
point at any time is given by
LẠ 02
Bt Ộxa Ở BỞ 1e)
Trang 2418 1.28 1.29 1.30 131
BOUNDARY VALUE PROBLEMS [CHAP 1 Write the boundary conditions in Problem 1.27 if (a) the ends ề=0 and x=T are insulated, : (b) the ends z=0 and Ủ=L radiate into the surrounding medium aceording to NewtonỖs law of cooling
The gravitational potential v at any point (x,y,z) outside of a mass m located at the point (X, Y, Z) is defined as the mass m divided by the distance of the point (x,y,z) from (X,Y,Z) Show that
Ừ satisfies LaplaceỖs equation V2v = 0
Extend the result of Problem 1.29 to a solid body
A string has its ends fixed at ề=0 and x=L It is displaced a distance h at its midpoint and then released Formulate a boundary value problem for the displacement y(x,t) of any point x
of the string at time ằ
CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS
1.32, Determine whether each of the following partial differential equations is linear or nonlinear, state the order of each equation, and name the dependent and independent variables @ Tatar tập =0 OO = aR O te a8 @ w+ St = +o @) Se 42h = 1.38 Classify each of the following equations as elliptic, hyperbolic or parabolic @) S$ 28 = 9 @) @2Ở1) 55 + #y cm + PT Động 6 + Oe = = ty (c) He Ở oft + 28% = a+ 8y (f) que Ở1) Se Ở Su = 0, M>0 (d) we + ay oe + ye = 0
SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS
134 Show that z(x,y) = 4eỞ-3* cos3y is a solution to the boundary value problem eo si 0, 'z(z,z/2) = 0, z(z,0) = 4eỞ3 1.35 (a) Show that %(z,) = xF(2xz+w) is a general solution of Ở Bây = 1.36 1.37 (6) Find a particular solution satisfying v(1,y) = y? `" ~
Find a partial differential equation having general solution +Ủ% 7= F{Ủ Ở 3y) + G(2z + 0) Find a partial differential equation having general solution
(a) z = etf(2yỞ 3x), (6) z = f(@u+y) + g(x Ở 2y)
GENERAL SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS 1.38 (a) Solve ề da Oy tay = 0 02z az
Trang 25CHAP 1] BOUNDARY VALUE PROBLEMS 19 1.39 1.40 1.41 1.42 Find general solutions of each of the following 02, 32 au ate , au au _ Pu, Fu _ (a) axe = 2w Ộ (b) ja + ay = 38u (e) ax2 ay? 0 82z 82z 83z 82z 83z 92z ỞỞ Ở Ở 8ỞỞ_ = Ở Ở Ở = 0
(4) Oa? Ou ay dy? 0 (e) 8x2 23x oy + ay? Find general solutions of each of the following
ou ou _ atu atu (a) an + ay = # (c) ant + 2 Say ay
ey _ 32 2 Be Ở az ae = i
(b) a OB + 12t (d) ax? dc ay + 2 x siny atu dtu atu
Solve aad + 3x5 ay> + yt Ở 16 : 822, 200 _ 10% _ F(rỞct) + Gứ + cĐ) Show that a general solution of m5 + rủ Gee 8 i + SEPARATION OF VARIABLES 1.43 1.41 1.45 1.46 1.47
Solve each of the following boundary value problems by the method of separation of variables
(a) aoe + ast = 0, Ủ0) = 4e-t ồ
\
(by #2 = at ty u(a,0) = 8e-5 3# + 2e-3e
(@\ Ủh = 4ÊỘ ot 322! Ấ(0,9 = 0, , , wắm,Đ = 0, ule,0) = 2sin8xỞ4sin5e +, , , ou _ du _ _ _ Bre Ở_ 9rx (d) aE ame? wẤ(0,É) = 0, - (2,1 = 0, u(Ủ,0) = 8 cosỞ 6 cos~Ở ou ou Ở = _ = Ở2 (e) 3 3 an? u(x, 0) 8eỞ22 ou Ou ou _ Ở = Ởz Ở GẹỞ4 (f) aE ax 2u, u(x, 0) 10e-* Ở 6eỞ4
gy %# = ử at 2x5? wWO,t) ; = 0, u(4,9 Ỉ , = 0, x@,0) = Ú, , = 6sinT Ộ+ 8i = 6ãn 3 sin 7%
Solve and give a physical interpretation to the boundary value problem ay _ =, ay Ở
if (a) f(x) = 5 sinwa, (b) f(a) = 3 sin 27x Ở 2 sin Bre
Ở ở _ 02w ` Solve 2 Ont 2u if u(0,t) = 0, (3,0 = 0, u(x,0) = 2 sinrz Ở sìn 4r#
Suppose that in Problem 1.24 we have y(x,0) = f(z), where 0<Ợ<2 Show how the problem can be solved if we know how to expand f(x) in a series of sines
Trang 26
Chapter 2 Fourier Series and Applications
THE NEED FOR FOURIER SERIES
In Problem 1.25, page 17, we saw that to obtain a solution to a particular boundary value problem we should need to know how to expand a function into a trigonometric series In this chapter we shall investigate the theory of such series and shall use the theory to solve many boundary value problems
Since each term of the trigonometric series considered in Problem 1.25 is periodic, it is clear that if we are to expand functions in such series, the functions should also be periodic We therefore turn now to the consideration of periodic functions
PERIODIC FUNCTIONS
A function f(x) is said to have a period P or to be periodic with period P if for all a, f(2+P) = f(x), where P is a positive constant The least value of P > 0 is called the least
period or simply the period of f(z)
Example 1
The function sin x has periods 27,47, 67, , since sin (x + 27), sin(x + 4z), sỉn ( + 6z), all equal sina However, 27 is the least period or the period of sin x Example 2 The period of sinnz or cosnx, where n is a positive integer, is 27/n Example 3 The period of tan # is 7 Example 4
A constant has any positive number as a period
Trang 27CHAP 2] : FOURIER SERIES AND APPLICATIONS 21
PIECEWISE CONTINUOUS FUNCTIONS f(a)
A function f(x) is said to be piecewise con- ;
tinuous in an interval if (i) the interval can be |
divided into a finite number of subintervals in |
each of which f(x) is continuous and (ii) the K zỞ0)
limits of f(x) as x approaches the endpoints of \ each subinterval are finite Another way of Ầ stating this is to say that a piecewise continu- Fa of ous function is one that has at most a finite
number of finite discontinuities An example of a piecewise continuous function is shown in
Fig 2-2 The functions of Fig 2-1(a) and (c)
are piecewise continuous The function of Fig
2-1(b) is continuous, Fig 2-2
The limit of f(x) from the right or the right-hand limit of f(x) is often denoted by
lim f +) = f@+0), where c>0 Similarly, the limit of f(x) from the left or the left- hand limit of f(x) is denoted by lim f(xỞề) = f(xỞ0), where ề>0 The values f(z +0)
en0
and f(xỞ0) at the point x in Fig 2-2 are asỔindicated The fact that ề>0 and ề>0
is sometimes indicated briefly by ề> 0+ Thus, for example, lim f(x+.ề) = f(x+0), lim f(xỞ.) = f(xỞ0) er 0+ es Ot 1 I i ị I | } 1 i | S I Ị Ị l } q8 ỞỞỞỞ~-
DEFINITION OF FOURIER SERIES
Let f(x) be defined in the interval (ỞL,L) and determined outside of this interval by
f(x +2L) = f(x), ie assume that f(x) has the period 2L The Fourier series or Fourier ex-
pansion corresponding to f(x) is defined to be
đọ K tn# Nae
9 + = (a cos + bn sin Ộ| (1) where the Fourier coefficients a, and bẤ are
L
|ề = if f(a) cos + dx
" Narr
E = ifn f(a ) sin Ở-Ở da Motivation for this definition is supplied in Problem 2.4
If f(x) mas the period 2L, the coefficients a, and bn can be determined equivalently from n= 0,1,2, (2) _ 1 e+2L trưa a = iJ f(x) cos > dx er Dn = iS x) sinỘ dx where c is any real number In the special case c= ỞL, (8) becomes (2) Note that the _?=0,1,2, (2) L constant term in (1) is equal to go _ 1 - 2 2L.) r f(x) dz, which is the mean of f(x) over a period
lf E=Ấ, the series (1) and the coefficients (2) or (2) are particularly simple ỘThe function in this case has the period 2z
It should be emphasized that the series (1) is only the series which corresponds to f(x)
Trang 28con-22 FOURIER SERIES AND APPLICATIONS [CHAP 2
verges to f(z) This problem of convergence was examined by Dirichlet, who developed conditions for convergence of Fourier series which we now consider
DIRICHLET CONDITIONS
Theorem 2-1: Suppose that
(i) f(x) is defined and single-valued except possibly at a finite number of points in (ỞL, L)
(ii) f(a) is periodic with period 2L
(iii) f(z) and f(x) are piecewise continuous in (ỞL, L)
Then the series (1) with coefficients (2) or (2) converges to (a) f(x) if x is a point of continuity
(0) fer + fe if x is a point of discontinuity For a proof see Problems 2.18Ở2.23
According to this result we can write
THX
f(t) = s + > (a cos TTỢ + bẤsin 2) (4)
n=1
at any point of continuity x However, if 2 is a point of discontinuity, then the 4eft side is replaced by 3[f(x +0) + f(x Ở0)], so that the series converges to the mean value of f(x +0)
and f(x Ở 0)
The conditions (i), (ii) and (iii) imposed on f(x) are sufficient but not necessary, i.e if
the conditions are satisfied the convergence is guaranteed However, if they are not satis-
fied the series may or may not converge The conditions above are generally satisfied in
cases which arise in science er engineering
There are at present no known necessary and sufficient conditions for convergence of
Fourier series It is of interest that continuity of f(x) does not alone insure convergence of a Fourier series
ODD AND EVEN FUNCTIONS
A function f(x) is called odd if f(-Ởx)=Ởf(v) Thus 2, xồ>Ở3x3+2z, sing, tan3x are odd functions
A function f(z) is called even if Ặ(=z) =Ặ(#) Thus 2Ổ, 22ồỞ42?+5, cosa, e*+e7* are even functions
The functions portrayed graphically in Fig 2-1(a) and 2-1(b) are odd and even respec-
tively, but that of Fig 2-1(c) is neither odd nor even
In the Fourier series corresponding to an odd function, only sine terms can be present In the Fourier series corresponding to an even function, only cosine terms (and possibly a
constant, which we shall consider to be a cosine term) can be present HALF-RANGE FOURIER SINE OR COSINE SERIES
A half-range Fourier sine or cosine series is a series in which only sine terms or only
Trang 29CHAP 2] FOURIER SERIES AND APPLICATIONS 25
function is desired, the function is generally defined in the interval (0,L) [which is half of the interval (ỞL,L), thus accounting for the name half-range| and then the function is specified as odd or even, so that it is clearly defined in the other half of the interval, namely
(ỞL,0) In such case, we have L dr = 0, bạ = zf f(x) sin TT đa for half-range sine series 0 (5) L bn = 0, Qa = if f(x) cos TT dư for half-range cosine series 0 PARSEVALỖS IDENTITY states that 2 LS, foyae = + Ế (+9) (6 if a, and b, are the Fourier coefficients corresponding to f(x) and if f(x) satisfies the Dirichlet conditions UNIFORM CONVERGENCE Suppose that we have an infinite series >) u(x) We define the Rth partial sum of the n=1 : series to be the sum of the first Fk terms of the series, i.e Ỗ R S.a) = > u,(2) (7)
Now by definition the infinite series is said to converge to f(x) in some interval if given any positive number ề, there exists for each x in the interval a positive number N such that
|S,(z) Ởf(x)| < ề whenever R>N (8) The number N depends in general not only on ề but also on x We call f(x) the sum of the series
An important case occurs when N depends on ề but not on the value of x in the interval
In such case we say that the series converges uniformly or is uniformly convergent to f(x)
Two very important properties of uniformly convergent series are summarized in the
following two theorems ố
Theorem 2-2: If each term of an infinite series is continuous in an interval (a,b) and the series is uniformly convergent to the sum f(x) in this interval, then
1 f(x) is also continuous in the interval
2 the series can be integrated term by term, i.e
S{z ua) dz = Ế Sf a(%) da (9)
Theorem 2-3: If each term of an infinite series has a derivative and the series of deriva- tives is uniformly convergent, then the series can be differentiated term byỖ term, i.e
? ` (10)
There are various ways of proving the uniform convergence of a series The most
obvious way is to actually find the sum S,(a) in closed form and then apply the definition directly A second and most powerful way is to use a theorem called the Weierstrass M
Trang 3024 FOURIER SERIES AND APPLICATIONS [GHAP 2
Theorem 2-4 (Weierstrass M test): If there exists a set of constants M,, n= 1,2, such that for all z in an interval |ua(x)|=M,, and if furthermore 2, M, Ủ n=1
converges, then >) wnỪ(x) converges uniformly in the interval Incidently,
n=1
the series is also absolutely convergent, i.e Ế |#a(z)| converges, under these
conditions
Example 5 we sin nz m
The series = converges uniformly in the interval (Ở7,7) [or, in fact, in any interval], since a set of constants Mẹ = ae can be found such that
: 2 1 sinnz| Ở 1
ỞỞỞI n2 | = SG mà and 2,2 g converges
INTEGRATION AND DIFFERENTIATION OF FOURIER SERIES
Integration and differentiation of Fourier series can be justified by using Theorems 2-2 and 2-3, which hold for series in general It must be emphasized, however, that those theorems provide sufficient conditions and are not necessary The following theorem for
integration is especially useful
Theorem 2-5: The Fourier series corresponding to f(z) may be integrated term by term from a to x, and the resulting series will converge uniformly,to Ặ F(u) du,
provided that f(x) is piecewise continuous in ỞL = 2 =L and both a and x are in this interval
COMPLEX NOTATION FOR FOURIER SERIES
Using EulerỖs identities,
eệ = cos + ?sin 6, e* = cosé@ Ở tsiné (11)
where i is the imaginary unit such that 1? = Ở1, the Fourier series for f(x) can be written in complex form as
f(2) = > Cụ einnr/L (12)
| , sẽ 7
where Cn = oF Sf f(Ủ)e~ mmz!L dẤ (12)
In writing the equality (12), we are supposing that the Dirichlet conditions are satisfied
and further that f(x) is continuous at x If f(x) is discontinuous at xz, the left side of (12)
should be replaced by Eto tie 9),
DOUBLE FOURIER SERIES
The idea of a Fourier series expansion for a function of a single variable x can be ex- tended to the case of functions of two variables x and y, i.e f(x,y) For example, we can
expand f(x,y) into a double Fourier sine series
f(z,y) = > Ế Ban sin" sin7# (14)
= ry
Trang 31CHAP 2] FOURIER SERIES AND APPLICATIONS 25
Similar results can be obtained for cosine series or for series having both sines and cosines These ideas can be generalized to triple Fourier series, etc
APPLICATIONS OF FOURIER SERIES
There are numerous applications of Fourier series to solutions of boundary value prob- lems For example:
1 Heat flow See Problems 2.25-2.29
2 LaplaceỖs equation See Problems 2.30, 2.31 3 Vibrating systems See Problems 2.32, 2.33 Solved Problems FOURIER SERIES 2.1 Graph each of the following functions B 0<z<5 = Period = 10 (a) f(x) La -g<z<0 eriod f(x) | -eỞ Period Ở*> + 3 T T T_T T TY T TỞỞ T T a Ở25 20 Ở15 10 ~5 0| 3 5 10 15 20 25 ỞỞỞỞ ỞỞỞỞ ỞỞ ầ ~ỞỞ~ -ÝỞỞ- ỞỞ~- Fig 2-3
Trang 3226 FOURIER SERIES AND APPLICATIONS [GHAP 2 0 0<=z<2 (c) f(x) = (1 3Sz<4 Period=6 0_ 4<Ộz<6 Ặ) [*Ở Period ỞỞ> ; _Ở _ỞỞ ỘỞỞ ( Ở Ở_ỞỞ _Ở 1 ny ~t Ởể ae ể en ể # -12 Ở10 Ở8 Ở8 ể4 _Ở8 0 2 4 6 8 10 12 14 Fig 2-5 Refer to Fig 2-5 above Note that f(x) is defined for all x and is discontinuous at a = +2,+4,+8,+10,+14, L L , 2.2 Prove f sin ỘỎ@ dy = f cos Ộ2 de =0 ỞL -L if k=1,2,3, L L _ kee, = iL kmg L _ _L Lo _ J, sin ~r de = ka ồồ a ig ẹồ8 ka + kz cos (Ởkz) 0 + knw _ L km L _ ba, bay _ S- cos = oe = 7, sin > 1Ợ Ole sin kz ler sin(-kr) = 0 L trữ Từn# L Hữn# Tnw [ 0 MAN A 2.3 Prove (a) (a f cos Ở=Ở cos Ở-Ở-dx = f sin ỞỞ sỉn Ởởd+ = -L L L -L L L \ Ù m=" L Mae gững b f sin Ở-dx = 0 (6) 1 T "
where m and ? can assume any of the values 1,2,8,
(a) From trigonometry:
cosA cosB = 4{cos(A Ở B) + cos (A + B)}, sinA sinB = 4{eos(A Ở B) Ở cos(A + B)} Then, if ?w z2 ụ, we have by Problem 2.2, L marx nee _ (mỞ m)rz = (m + n)rz _ J, C08 ỞTỞ Ạ0ậ ~TỞ dx = NI es + a dx = 0 Similarly, if m #1,
L mare nie _ 1 L (mỞn)re _ (m+ n)ra _ J sin _ sin dy = a {eos a cos dx = 0
If m=n, we have
L mare nae _ 1 L 2nrx _ Sos T1 C05 dz = 2S, (1 + cos L ) ae = OL
L Thư Nae _ it L _ 2nrz _
J, sin T sin L dz = oJ ụ cos ) ae = L
Note that if mỞ=xn-=0 these integrals are equal to 2L and 0 respectively
(6) We have sinA cosB = f{sin (AỞB)+sin(A+B)} Then by Problem 2.2, if mn,
L
Trang 33CHAP 2] FOURIER SERIES AND APPLICATIONS Ta 27
2.4
3.5
If m=n , L tư nee 4 Ở in T 2 "7# dy = Q
J sin a cos a 3 Soe
The results of parts (a) and (b) remain valid when the limits of integration ỞL, L are replaced by c,ằ+2L respectively
If the series A + Ế (a cos T = + ba sin +) converges uniformly to f(x) in (ỞL, L),
n=1
show that for 1 =1,2,3, , 4
(a) Gn = LÍ, f(x) cos TT da, (6) bạ = LÝ, f(x) sin Ộ7 da, (ce) A= ae
~ ne
(a) Multiplying Ặf@ = A + 5 ứ cos Ộ=~ n=1 T- + b, sin =} (1)
by cos TT and integrating from ỞL to L, using Problem 2.8, we have lì A Ặ L cos mae dx ồ L +3 {en f cosỘ F* cos ỘF= da + by f cos MEE sin EE do | n=1 ểã = a,L if mz 0 (2) L Ặ Ặ(Ủ) cos dx -L + L Thus Gm = LÍ f(a) cos TTỢ dx if m =1,2,3, ỞL (b) Multiplying (1) by sin Ộ2Ợ and integrating from ỞL to L, using Problem 2.3, we have + L Mare L Mire S f(z) sin L d A 4, sin dy oo i + 3S fon f sin TTỢ Ộ cos ỘTỢ da + by sin mE in = a n=1 -L H = b,,L Ở 1 1 Mare Ở Thus by == L f(%) sin dx if m= 1,2,3, _ L (c) Integration of (1) from ỞL to L, using Problem 2.2, gives L , L Ặ fe)de = 2AL L or A = & Ặ flee) dex 2L _ ỞL 1 (Ủ đọ Putting m=0 in the result of part (a), we find a) = +f f(z)dx andso A = 2 ỞE
The above results also hold when the integration limits ỞLZ, L are replaced by ec, e+ 2L Note that in all parts above, interchange of summation and integration is valid because the series is assumed to converge uniformly to f(x) in (-L,Z) Even when this assumption is not warranted, the coefficients a, and b,, as obtained above are called Fourier coefficients corresponding
to f(x), and the corresponding series with these values of a,, and đẤ is called the Fourier series
corresponding to f(x) An important problem in this case is to investigate conditions under which this series actually converges to f(x) Sufficient conditions for this convergence are the Dirichlet conditions established below in Problems 2.18-2.23
(a) Find the Fourier coefficients corresponding to the function
f(a) 0 -ã<z<0 Period 10
x = 1 =
Trang 3428 FOURIER SERIES AND APPLICATIONS (CHAP 2
(b) Write the corresponding Fourier series
(c) How should f(x) be defined at ề= Ở5, = 0 and ề=5 in order that the Fourier
series will converge to f(z) for -5 S25?
The graph of f(x) is shown in Fig 2-6 below f (x), ỞỞ Period => Ỉ _ỞỞỞ FỞ- ~_ỞỞ_~ lỗ ~10 ỞẾ Fig 2-6 (a) Period = 9L = 10 and L=5 Choose the interval ằ to c+ 2L as Ở5 to 5, so that ằ = Ở5 Then c+2L "số ể.h K e _ 1 0 ane 4 Ƒ Trưm _ 8 { Trữ = HS, (0) cos ~~ dx + Jo (3) cos = de = 3J, cos ~~ dx 0 if n#0 = 3/5 5, nex\! ~ < 5 0 Orx 3 ồồ _ If n=0, a, = a = 3 cost dz = 8 fd = 8, 1 or 2b ,_ T# _ỷ 1 f H6 | by = T1 Ặ f(x) sin TT dx = 5 f(x) sin EỞ dx ồ _ 8 > nxn = ậ if 0) sinỘ dy + f (8) s5 dự = 3S, sin TEỞ dư _ 383 5 mzz\|Ế _ 81 Ở cos nz) = =| -Ở cosỞ = 5 Tim B 0 nr (6) The corresponding Fourier series is đọ 5 + = (a, cos TT < 1i + 6b, sinỢ L woe 2) = _ 3 = 3(1 Ở cos nr) 9 + 2 Tap sin = Tư
_Ò 3 , 6f rH 1 3ưz 1 đm& su
Ở 3 8 (sin + din nop + gsinỘg + )
(c) Since f(x) satisfies the Dirichlet conditions, we can say that the series converges to f(x) at all f(x +0) + f(x Ở0)
2
points of continuity and to at points of discontinuity At ề=-Ở5, 0 and 5, which are points of discontinuity, the series converges to (3+ 0)/2 = 3/2, as seen from the graph The series will converge to f(x) for ỞB x5 if we redefine f(x) as follows: 3/2 # =: ỞB 0 ~Bca <0 f(z) = 3/2 x= 0 Period = 10 3 0<xz<5đ 3/2 =Ủ=Ở=5
2.6 Expand Ặ(z) = z2, 0< z< 9z, in a Fourier series if the period is 2z
Trang 35FOURIER SERIES AND APPLICATIONS 29 CHAP 2] f(x) / / / / / / ⁄ ⁄ /ứ ⁄ ⁄ ⁄ : ⁄ 4ồ ⁄ ⁄ ⁄ ⁄ ⁄ | ⁄ z ⁄ ~ ⁄ ~ ⁄ _~Z ⁄ Ỷ _ỘZ ⁄ ~ x | I T O Ị ĩ T Ở6r ỞÁz 4u | 2z 4z 6m Fig 2-7 Period = 2L = 27 and L=r Choosing c = 0, we have _ 1 é+2L nee _ 1 Ặ 2 om = F Ặ F(x) cos da = J, x? cos nx dx - 1 (02) sin nx Ở (8z) Ởeos se 42 Ở sin nx \| |?" _ 4 nầ0 7 +: r3 0 neỖ 1 are If n=0, @ == A (Ved: = ĐỂ 1 ồr Ure 1y" oe _ 1 mre ay =D d by, L Ặ f(x) sin T dx 7 Jo x sinna dx 2m7 Ở = te = Z (3z (- sin ne) + (S872 sone) Ở a 7 n 0 Ề = Ar Then f(z) = w = + 3 đã S, cosa Ở $F sin nz for 0 <a < 27 : 1 1 1 a
2.7 Using the results of Problem 2.6, prove that pt 5 + 3 tree = 6ồ
At ề=0 the Fourier series of Problem 2.6 reduces to = + Ế +
n=1
But by the Dirichlet conditions, the series converges at ề =0 to 3(0 + 4z?) Ở 2m3, Hence the desired result
ODD AND EVEN FUNCTIONS HALF-RANGE FOURIER SERIES
Trang 36FOURIER SERIES AND APPLICATIONS [CHAP 2 cosx <#ụ<z (b) f(x) = 0 zr < # < 2m Period = 27 From Fig 2-9 below it is seen that the function is neither even nor odd F(x) 7S, 1 7s, ể ` ` ` x ể ` T T T N TỢ ỘoO O Xo 2m \ 82 ` Se ` NGẰ Fig 2-9 (e) Ặ(z)=z(10Ởz), 0<z<10, Period = 10 From Fig 2-10 below the function is seen to be even F(z) mm ⁄ ` 7 ⁄ ` ` \ / ⁄ ` ⁄ \ 25 / \ ⁄ \ ⁄ Ns \ | / Vv # Ẩ | f 10 0 5 10 Fig 2-10 2.9 Show that an even function can have no sine terms in its Fourier expansion Method 1 No sine terms appear if 6, =0, n=1,2,3, To show this, let us write 1ặ _ Tư _Ở tl ồ nie L cẬ
LÍ, f(x) sỉn da = is, f(x) sin dx += iS, f(x) sin T dx (1) If we make the transformation #ề = Ởw in the first integral on the right of (2), we obtain
0
* 1L
1Í, f(x) sin dx if Ặ(Ởw) sin( 25) du Ở= Ộặ9 f(Ởu) sin "2 du
= NÃN fu) sin du = -2f f(x) sin" dư (2)
where we have used the fact that for an even function f(Ởu) = f(u) and in the last step that the dummy variable of integration u can be replaced by any other symbol, in particular x
Trang 37CHAP 2] FOURIER SERIES AND APPLICATIONS 31
đa ~ ưa
and so 2 by sin TTỢ = 0, ie, f(z) = a + > an cos
and no sine terms appear This method is weaker than Method i since convergence is assumed
In a similar manner we can show that an odd function has no cosine terms (or constant term)
in its Fourier expansion : - 2 ch Trữ _ 2.10 If f(z) is even, show that (a) On =F Ặ f(a) cos 5" dz, (b) bẤ=0 - ềT0 Ổ (a) a, = if, f(x) cos ỘTỢ đứ = an F(x) cos du + = if f(x) cos ỘEỢ dư Letting 2 = Ởu, 0 L _ L
ty Ấ7ệ cos Ộ= dz = tJ f(Ởw) cos (=#") du = ; - 3 Ặ fw) cos , du
since by definition of an even function f(Ởu) = f(u) Then
L L L
a, = iJ, flu) cos ỎE% du + Ff f(x) cos da = zs, f(a) cos ỘTTỢ dư (b) This follows by Method 1 of Problem 2.9
2.11 Expand Ặ(z) = sinz, 0< # < 7a, ina Fourier cosine series
A Fourier series consisting of cosine terms alone is obtained only for an even function Hence
we extend the definition of f(a) so that it becomes even (dashed part of Fig 2-11) With this
extension, f(x) is defined in an interval of length 27 Taking the period as 27, we have 2L = 2z, so that L= 7 F(x) ỘTN i oT ` Ny 7 \ \ 7 ⁄ ` \ ⁄ ⁄ ` Ổ4 ⁄ V2 NV # Tt | J Ĩ Ở2z Ởz 7 2m Fig 2-11 By Problem 2.10, 6, =0 and wh, + a, = = Í f(x) cosỘTỢ dx = SỆ Sỉn # eỦos 2z dx 70 7/9 1T Ở Tv
Trang 3832 FOURIER SERIES AND APPLICATIONS [GHAP 2 9 2 & (1 + cos nz)
Then F(a) = 7 Ở 7 2 ei cos NX
_ 2 4f cos2x cos 4a cos6% , - La A( geste 4 cose 4, O86 4 )
2.12 Expand f(z)=ề, 0<z<2, in a half-range (a) sine series, (b) cosine series
(a) Extend the definition of the given function to that of the odd function of period 4 shown in Fig 2-12 below This is sometimes called the odd extension of f(x) Then 2L=4, L=2 f(x) ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ O 7 # T zl i 7 Ẩ A I - z1 HẮ 2 Z4 6 ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ Fig 2-12 Thus a, =0 and _ 2 L ,_ Tư _ 3f 2 nwt b, = 7 f, f(x) sin-TỞ dx = 5 Ặ # sin~5 dx _ Ở2 nee \ _ =4 ụmz\||J _ Ở4 = C cos 15) ` sin ) COS Tr
Then fz) = n=1 "7 3 =4 cos nr sin 2 =
= Af int Ở 1 gin 27Ợ 4 1 gi 87% _
T 2 2 2 3 2
(b) Extend the definition of f(a) to that of the even function of period 4 shown in Fig 2-13 below
Trang 39GHAP 2] FOURIER SERIES AND APPLICATIONS 33 = 4 nee 1+ = ays (C08 Mar Ở1) cos =~ Then f(x) - 8/ ett 4 1 pg Stt y Lg bee = "Ộ + gy cos + Fe C085 + )
It should be noted that although both series of (a) and (6) represent f(x) in the interval 0 < ề < 2, the second series converges more rapidly
PARSEVALỖS IDENTITY
2.13 Assuming that the Fourier series corresponding to f(x) converges uniformly to /(#)
in (-L, L), prove ParsevalỖs identity
1 x 2 độ 2 2
zs (Kade = sở + Đ (+ b2)
Ởk n=1
where the integral is assumed to exist
If f(z) = 3 + 3 C cos 7Ợ + 6b, s2) , then multiplying by f(x) and integrating n=1 term by term from ỞL to L (which is justified since the series is uniformly convergent), we obtain L a, L = L L f -L {f(a)}2daz > = _Ở Ặ -Ặ(@Ủ) de + ứ 2) L n=1 Ja, Ặ -L f(x) cos nae ae + db, Ặ L ỞE f(a) sin nee dx TL đã Ủ = sL+LẾ (ậ+b) (1) =1 where we have used the results L L L
Ặ _ Hl) cos 2% de = Lay, Ặ | Hla) sin" de = Ldy Ặ _ Weds = Lay (2)
obtained from the Fourier coefficients
The required result follows on dividing both sides of (7) by L ParsevalỖs identity is valid under less restrictive conditions than imposed here In Chapter 3 we shall discuss the significance
of ParsevalỖs identity in connection with generalizations of Fourier series known as orthonormal
series
2.14, (a) Write ParsevalỖs identity corresponding to the Fourier series of Problem 2.12(0)
(b) Determine from (a) the sum S of the series at x + a +:-:+ = tree,
(a) Here L = 2; ag=2; a, = =ặ; (cos nz Ở1),n +0; b,=0 Then ParsevalỖs identity becomes
2 2 20
af, Wenrde = 50 ae = Ba 2 S18 corns 1 1)
o 3 = 2+ She gt ht), ie Stet ate ==
() s=i+h+h+ 14 24 34 = (i+h+kt 14ồ 347 54 )+ tátá+ 241 ` 4+ ` 64
= (Ãrátkth) tá(rá rất)
_ 7t 8: : _ wt
Trang 4034 FOURIER SERIES AND APPLICATIONS {CHAP 2
2.15 Prove that for all positive integers M, as 3 1 L (@+b) = = f(z))? da + > J Ư@)) where a, and b, are the Fourier coefficients corresponding to f(x), and f(x) is assumed piecewise continuous in (ỞL, L) đo s ATL `
Let Sule) = 3+ = (4 cos x~ = + b, sin k2 (1)
- For M = 1,2,8, this is the sequence of partial sums of the Fourier series corresponding to f(z) L We have f {f(z) ỞSy(a)}2dx 2 0 (2) -L since the integrand is non-negative Expanding the integrand, we obtain L f _ ia}? de (8) IIA L L
2 Ặ f(a) Syg(ae) dae Ở f : S(z) dư
Multiplying both sides of (1) by 2f(x) and integrating from ỞL to L, using equations (2) of Problem 2.13, gives L az M 2 ĐỆ /@)Sw@)dz = 2+2 + S (ak + 0) ỞL n=1 4) Also, squaring (1) and integrating from ỞL to L, using Problem 2.3, we find L Q aM Ặ Su(x)de = L+Ở+ 3 (a2+ 0?) (5) ~L 2 Ấ=i
Substitution of (4) and (5) into (2) and dividing by L yields the required result Taking the limit as M > ẹ, we obtain BesselỖs inequality
2 wo L
ots a+ s tf _ ila de (6)
If the equality holds, we have ParsevalỖs identity (Problem 2.13)
We can think of S),(x) as representing an approximation to f(x), while the left hand side of (2), divided by 2L, represents the mean square error of the approximation ParsevalỖs identity indicates that as M > ẹ the mean square error approaches zero, while BesselỖs inequality indicates the possibility that this mean square error does not approach zero
The results are connected with the idea of completeness If, for example, we were to leave out one or more terms in a Fourier series (cos 47Ợ/L, say), we could never get the mean square error to approach zero, no.matter how many terms we took We shall return to these ideas from a gen- eralized viewpoint in Chapter 3
INTEGRATION AND DIFFERENTIATION OF FOURIER SERIES _
2.16 (a) Find a Fourier series for Ặ(2) =2, 0<z<2, by integrating the series of Problem 2.12(a) (b) Use (a) to evaluate the series > C1 ` m8 Ở